# Lower bound for $\sum_{i=1}^n x_i f(x_i)$ when $\sum_{i=1}^{n}x_i = K$

Considering,

1. the set of all n dim. vectors $\{x_i\}_{i=1,...,n}$ such that $x_i \geq 0$ and $\sum_{i=1}^{n}x_i = K$
2. Any continuous and strictly increasing function $f^+(x)$ : $\mathbb R^+ \to \mathbb R^+$ (i.e. where $f^+(x+\epsilon)> f^+(x) \quad \forall \epsilon >0 \quad x \in \mathbb R^+$)

My question: Is it true in general that $\sum_{i=1}^n x_if(x_i)\geq Kf(\frac{K}{n})$ for any $\{x_i\}$ respecting 1. and $f^+$ ?

Using cauchy-schwartz ineq. I was able to prove this result for linear functions $f^+(x)=ax+b, \quad a,b>0$ , then:

$\sum_{i=1}^n x_if(x_i)=Kb+a\sum_{i=1}^n x_i^2\geq Kb+\frac{aK^2}{n}$ since $n\sum_{i=1}^n x_i^2\geq (\sum_{i=1}^n x_i)^2=K^2$ .

(where $n\sum_{i=1}^n x_i^2=K^2$ if $x_i=K/n$)

Intuitively I was thinking that this should lead to a more general proof of the statement since for any interval $I=[x_{min},x_{max}]$, we can always find $a,b>0$ such that $ax+b\leq f^+(x) \quad \forall x \in I$. But however I didn't find yet the way to conclude using this path.

I am missing something? Is there another way to proceed?

Thank's for your help!

• By Jensen, this is immediate for convex $f$: $\sum x_i f(x_i) = K \sum \frac{x_i}{K} f(x_i) \ge K f(\sum x_i^2 / K)$, and since $f$ is monotone increasing and by C-S $\sum x_i^2 / K \ge (\sum x_i)^2 /(nK) = K/n$, we're done. – Ofir Gorodetsky Jul 4 '15 at 11:58

## 2 Answers

Actually,the conclusion is negative.

By chebyshev's theorem, we have $$\sum_{i=1}^{n}{x_{i}\cdot f(x_{i})}\geq \frac{\sum_{i}^{n}{f(x_{i} ) } }{n} \times \sum_{1}^{n}{x_{i} }$$ $$\forall x_{1}<x_{2}<...<x_{n}\in \left\{ x_{i} \right\} , f^{+}(x+\epsilon ) >f^{+}(x+ ) ,\forall \epsilon >0,x\in R^+$$

By Jensen, if $f(x)$ is convex, we have $$\frac{\sum_{i}^{n}{f(x_{i} ) } }{n}\times \sum_{1}^{n}{x_{i} } \geq f(\frac{\sum_{1}^{n}{x_{i} } }{n} )\times \sum_{1}^{n}{x_i}$$ $$\forall x_{1}<x_{2}<...<x_{n}\in \left\{ x_{i} \right\} , f^{+}(x+\epsilon ) >f^{+}(x+ ) ,\forall \epsilon >0,x\in R^+$$
So,if $f(x)$ is convex,we have $$\sum_{i=1}^{n}{x_{i}\cdot f(x_{i})}\ \geq f(\frac{\sum_{1}^{n}{x_{i} } }{n} )\times \sum_{1}^{n}{x_i}$$ (only Jensen is not enough to prove this)

By Jensen, this is true if $g(x):=xf(x)$ is convex:

$\sum x_i f(x_i) = \sum g(x_i) \ge n g(\sum x_i /n) = Kf(K/n)$.

As Jensen characterizes convex functions, if we find $f$ subject to condition (2) with $xf(x)$ non-convex, we obtain a counterexample.

Just take $f$ whose derivative is $f'(x) = \frac{1- \cos x}{x^2} \ge 0$: Calculating $(xf(x))''$, we find that it is $\frac{\sin x}{x}$, which is not always non-negative.